- #1

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that are the solutions of the equation [tex]15x^2+y^2=2^{2000}[/tex]

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- Thread starter Anzas
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- #1

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that are the solutions of the equation [tex]15x^2+y^2=2^{2000}[/tex]

- #2

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a = 2^2000 + 2^2000 * t,

b = -14*2^2000 - 15*2^2000 * t,

where t is some integer. Now, if x^2 = a and y^2 = b (with x, y natural), then it's necessary (but obviously not sufficient) that both a and b be positive. This places some severe restrictions on t, in fact, if you try to solve the system

a >= 0

b >= 0

you'll find that t = -1 is the only possibility, but then a = 0, which isn't odd.

- #3

Hurkyl

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For various reasons, the diophantine equation 15a + b = 2^2000 has the solutions

a = 2^2000 + 2^2000 * t,

b = -14*2^2000 - 15*2^2000 * t,

I'd be curious to know those reasons...

- #4

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- #5

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so i suppose the question remains open?

- #6

Zurtex

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Are you sure it has any solutions?

I've messed about with this and found if there is a solution then y must be one of the 4 forms:

[tex]y = 1 + 30c \quad \text{or} \quad y = 11 + 30c \quad \text{or} \quad y = 19 + 30c \quad \text{or} \quad y = 29 + 30c \quad c \in \mathbb{Z}[/tex]

I've messed about with this and found if there is a solution then y must be one of the 4 forms:

[tex]y = 1 + 30c \quad \text{or} \quad y = 11 + 30c \quad \text{or} \quad y = 19 + 30c \quad \text{or} \quad y = 29 + 30c \quad c \in \mathbb{Z}[/tex]

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